By tightness of \(\nu _0\) , we can choose a sequence of coercive functionals \(\Phi _i\)  such that
\(\int \Phi _i(p_i(x)) \,\mathrm {d}\nu _0 \le 2^{-i}, \qquad \mbox{for all $i\in \mathbb {N}$},\) 
and consider the functional \(\mathcal {A}(\eta ):C([0,T];\mathbb {R}^\mathbb {N})\rightarrow [0,+\infty ]\)  given by
\(\mathcal {A}(\eta ):=\left\lbrace \begin{aligned}&\sum _{i=1}^{\infty } \left(\Phi _i(p_i\circ \eta (0))+\int _0^T M\Psi \left(\frac{|\dot{\eta }(t)|_{\infty }}{M(1+|p_1\circ \eta (t)|)}\right) \mathrm {d}t \right) \quad & &\mbox{if $\eta \in AC_w([0,T];\mathbb {R}^\mathbb {N})$,} \\&+\infty \quad & &\mbox{otherwise.}\end{aligned} \right.\) 
It is clear that \(\mathcal {A}\)  is coercive in \(C([0,T];\mathbb {R}^\mathbb {N})\) , and its sublevel sets contain curves that are absolutely continuous with respect to \(|\cdot |_{\infty }\) . This follows from the fact that \(\sup _{t\in [0,T]}{|p_1\circ \eta |}\)  is bounded on the sublevel sets of the functional
\(\left(\Phi _1(p_1\circ \eta (0))+\int _0^T M\Psi \left(\frac{|(p_1\circ \eta )^{\prime }(t)|}{M(1+|p_1\circ \eta (t)|)}\right) \mathrm {d}t \right).\) 
Now, for every \(n\in \mathbb {N}\) , we define the marginals \(\mathcal {P}(\mathbb {R}^n)\ni \nu ^n_t:=(\pi _n)_{\#}\nu _t\)  and corresponding vector fields \(c_{t}^n: \mathbb {R}^n\rightarrow \mathbb {R}^n\)  by
\(p_i \circ c_{t}^n:=\frac{\mathrm {d}\, (\pi _n)_{\#}\left((p_i\circ c_t)\, \nu _t\right)}{\mathrm {d}\nu _t^n}.\) 
Note that \((\nu ^n,c_t^n)\)  satisfies the continuity equation in \(\mathbb {R}^n\) . By Jensen's inequality, and the fact that \(|z_1|\le |z|\le n |z|_{\infty }\) , for \(z=(x_1,\ldots ,x_n)\in \mathbb {R}^n\) , we have that
\(\begin{aligned}T \Psi \left(\frac{1}{n MT}\int _0^T \int \frac{|c_t|}{(1+|x_1|)}\, \mathrm {d}\nu _t \, \mathrm {d}t \right) &\le T \Psi \left(\int _0^T \int \frac{|c_t|_{\infty }}{MT(1+|x_1|)}\, \mathrm {d}\nu _t \, \mathrm {d}t \right)\\&\le \int _0^T \int \Psi \left(\frac{|c_t|_{\infty }}{M(1+|x_1|)}\right) \mathrm {d}\nu _t \, \mathrm {d}t.\end{aligned}\) 
and in particular
\( \int _0^T \int \frac{|c_t|}{(1+|x_1|)}\, \mathrm {d}\nu _t \, \mathrm {d}t < \infty .\) 
Hence, we can apply the finite-dimensional version of [1]}. Embedding this into \(\mathbb {R}^\mathbb {N}\) , we obtain the probability measure \(\lambda _n\)  over \(C([0,T],\mathbb {R}^\mathbb {N})\) , concentrated on absolutely continuous curves satisfying \(\dot{\gamma }=c_t^n(\gamma )\) , and such that \((e_t)_{\#}\lambda _t=\nu _t^n\) . We immediately see that
\(\sup _{n\in \mathbb {N}} \int \mathcal {A}(\gamma )\, \mathrm {d}\lambda _n(\gamma ) <\infty ,\) 
which yields the tightness of \(\lambda ^n\) .
Consider any converging sequence \(\lambda ^n\)  (up to renumbering) and its limit \(\lambda \in \mathcal {P}(C([0,T];\mathbb {R}^\mathbb {N}))\) . Since the sequence \((\nu _t^n)_{n\in \mathbb {N}}\)  clearly converges to \(\nu _t:=(e_t)_{\#}\lambda \)  in \(\mathcal {P}(\mathbb {R}^\mathbb {N})\)  for every \(t\in [0,T]\) , it remains to show that \(\lambda \)  is concentrated on solutions of \(\dot{\gamma }=c_t(\gamma )\) . In fact we will show that
\(\int \frac{\left|p_i \circ \gamma (t)-p_i\circ \gamma (s)-\int _s^t p_i\circ c_r(\gamma (r))\, \mathrm {d}r\right| }{1+ \Vert p_1\circ \gamma \Vert _{\infty }}\,\lambda (\mathrm {d}\gamma )=0\qquad \text{for each $i\in \mathbb {N}$ and any $0\le s\le t\le T$.}\) 
Note that it suffices to show that for any vector field \(d:[0,T]\times \mathbb {R}^\mathbb {N}\rightarrow \mathbb {R}\)  with \(d_t\)  being \(k\) -cylindrical for every \(t\in [0,T]\) , we have that
\(\int \frac{\left|p_i \circ \gamma (t)-p_i\circ \gamma (s)-\int _s^t d_r(\gamma (r))\, \mathrm {d}r\right| }{1+\Vert p_1\circ \gamma \Vert _{\infty }}\,\lambda (\mathrm {d}\gamma )\le \int _s^t\int _{\mathbb {R}^\mathbb {N}} \frac{|p_i\circ c_r-d_r|}{1+|p_1|} \,\mathrm {d}\nu _r\, \mathrm {d}r,\) 
since then we can use density of time-dependent cylindrical functions in \(L^1((1+|p_1|)^{-1} \nu _s \,\mathrm {d}s)\)  and the fact that for all \(s\)  it holds that \(|p_1\circ \gamma (s)|\le \Vert p_1\circ \gamma \Vert _{\infty }\) .
To prove (REF ), recall that \(\lambda ^n\)  is concentrated on absolutely continuous solutions of \(\dot{\gamma }_s=c^n_s(\gamma _s)\) . Hence,
\(\begin{aligned}\int \frac{\left|p_i \circ \gamma (t)-p_i\circ \gamma (0)-\int _0^t d_s(\gamma (s))\, \mathrm {d}s\right| }{1+\Vert p_1\circ \gamma \Vert _{\infty }}\lambda ^n(\mathrm {d}\gamma ) &= \int \frac{\left|\int _0^t(p_i\circ c^n_s(\gamma (s))-d_s(\gamma (s)))\, \mathrm {d}s\right| }{1+\Vert p_1\circ \gamma \Vert _{\infty }}\lambda ^n(\mathrm {d}\gamma )\\&\le \int \frac{\int _0^t |p_i\circ c^n_s-d_s|(\gamma (s))\, \mathrm {d}s }{1+|p_1(\gamma (s))|}\lambda ^n(\mathrm {d}\gamma )\\&\le \int _0^t \int _{\mathbb {R}^\mathbb {N}} \frac{|p_i\circ c^n_s-d_s|}{1+|p_1|}\, \mathrm {d}\nu ^n_s\, \mathrm {d}s\\\end{aligned}\) 
Note that the integrand on the left-hand side is continuous in \(\gamma \) . Therefore, since for \(n\ge k\)
\((p_i\circ c_s^n-d_s)\nu _s^n=(\pi _n)_{\#}((p_i\circ c_s-d_s)\nu _s),\) 
the result then follows after taking the limit \(n\rightarrow \infty \) .
